Suppose $f$ is a homomorphism from $A \to B$, and $C \subset A$. Further suppose that $r$ is a function that maps $C \to B$ $|$ $r(C) = f(C)$
and since I wish to prove that $r$ is also a homomorphism I would show that:
since $r$ behaves like $f$ but is simply restricted to the subgroup of $A$ (called $C$), and $f$ is a homomorphism by definition, if there exists $x,y\in C$ such that $f(x, y) = f(x) f(y)$ then it's also true that $r(x, y) = r(x) r(y)$ for some $x,y\in C$, thus showing that $r$ is a homomorphism
and if this is true, then it is additionally true that if we limit ker$(f)$ to the scope of the subset $r(K) ∀k\in K$, then ker$(r)$ = ker$(f) \cap k$
please let me know what your thoughts are on this problem and correct me where you see fit; all help and input is appreciated
If $r:=f_{|C}$, then $r(c_1c_2)=f(c_1c_2)=f(c_1)f(c_2)=r(c_1)r(c_2)$, and $r$ is indeed a homomorphism.
Then:
\begin{alignat}{1} \operatorname{ker}(r) &= \{c\in C\mid r(c)=e_B\} \\ &= \{c\in C\mid f(c)=e_B\} \\ &= \{c\in A\mid f(c)=e_B \wedge c\in C\} \\ &= \{c\in A\mid f(c)=e_B\}\cap \{c\in A\mid c\in C\} \\ &= \operatorname{ker}(f)\cap C \\ \end{alignat}